Introduction Taylor s Theorem Einstein s Theory Bachelier s Probability Law Brownian Motion Itô s Calculus. Itô s Calculus.

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1 Itô s Calculus Christopher Ting Christopher Ting : christopherting@smu.edu.sg : : LKCSB 5036 October 21, 2016 Christopher Ting QF 101 Week 10 October 21, /70

2 Table of Contents 1 Introduction 2 Taylor s Theorem 3 Einstein s Theory 4 Bachelier s Probability Law 5 Brownian Motion 6 Itô s Calculus Christopher Ting QF 101 Week 10 October 21, /70

3 Botanical Brownian Motion Brownian Motion: Experiment Brownian Motion: Animation What are the characteristics of Brownian motion? Source: Robert Brown Christopher Ting QF 101 Week 10 October 21, /70

4 Bachelier and Einstein They derived a scaling law for the rate of diffusion independently, one for market prices, and the other for molecules. Source: Louis Bachelier Source: Albert Einstein Christopher Ting QF 101 Week 10 October 21, /70

5 Taylor s Theorem On Taylor s Theorem:...le principal fondement du calcul différentiel... Joseph-Louis Lagrange ( ) Source: Biographie universelle des musiciens et bibliographie générale de la musique: Sa - Zy, Volume 8, Page 337. Taylor s Theorem A differentiable function may be represented by Taylor s series expansion of the function f(x) about a finite real number a: f(x) = f(a) + f (a)(x a) + f (a) (x a) 2 + 2! + f (n) (λ) (x a) n + n! Christopher Ting QF 101 Week 10 October 21, /70

6 The true and best way of learning any Art, is not to see a great many Examples done by another Person, but to possess ones self first of the Principles of it, and then to make them familiar, by exercising ones self in the Practice. Source: New Principles of Linear Perspective, Page vi Source: Brook Taylor Christopher Ting QF 101 Week 10 October 21, /70

7 A Link to Your Pre-U Math When a = 0, we have the Maclaurin series expansion! f(x) = f(0) + xf (0) + x2 2! f (0) + x n = n! f (n) (0) n=0 Two special cases are very useful in Quantitative Finance e x = 1 + x + x2 2! + = k=0 log(1 + x) = x x2 2 + x3 3 + = k=1 x k k! for all x ( 1) k+1 x k k for 1 < x 1 Christopher Ting QF 101 Week 10 October 21, /70

8 Geometric Intuition for Approximation a+h a f (x)dx = f(a + h) f(a) = f(a + h) = f(a) + The first-order Taylor series approximation of f(x) is the slope of the function at P, i.e. tan θ = f (a). The value of the function at Q, f(a + h), is approximated by the ordinate of R, f(a) + hf (a). a+h a f (x)dx (1) Christopher Ting QF 101 Week 10 October 21, /70

9 First-Order Approximation As a first-order approximation, if h is sufficiently close to a, we approximate f (a) as f (x). Then f(a + h) f(a) + hf (a). Noting that x a = h, we have f(x) f(a) + (x a)f (a). If f(x) is at least twice differentiable, then f (x) f (a) + (x a)f (a), Christopher Ting QF 101 Week 10 October 21, /70

10 Second-Order Approximation The second-order approximation is f(a + h) f(a) + f (a)h + f (a) h 2. 2 Proof: Substitute f (x) f (a) + (x a)f (a) into the integral in (1) f(a + h) f(a) + a+h a f(a) + f (a)x ( f (a) + (x a)f (a) ) dx a+h a f(a) + f (a)h f (a)h f (a)(x a) 2 a+h a Christopher Ting QF 101 Week 10 October 21, /70

11 A Proof of Taylor s Expansion by Pre-U Math Assumption: The function f(x) is expressible as a polynomial: f(x) = c 0 + c 1 (x a) + c 2 (x a) 2 + c 3 (x a) 3 +, where a is a constant, and we want to find c i for i = 0, 1,...,. To find c 0, set x = a. Then all x a terms become zero. Accordingly, c 0 = f(a). To find c 1, take the derivative of f(x), resulting in f (x) = c 1 + 2c 2 (x a) + 3c 3 (x a) 2 + Again, set x = a. All x a terms vanish and we have c 1 = f (a). Christopher Ting QF 101 Week 10 October 21, /70

12 A Proof of Taylor s Expansion by Pre-U Math (Cont d) To obtain c 2, take the second derivative: With x = a, we obtain f (x) = 2c c 3 (x a) + c 2 = 1 2 f (a). To obtain c 3, take the third derivative: f (x) = 3 2c c 4 (x a) + With x = a, we have c 3 = f (a). Christopher Ting QF 101 Week 10 October 21, /70

13 A Proof of Taylor s Expansion by Pre-U Math (Cont d) In general f (n) (x) = n! c n + (n + 1) n! c n+1 (x a) +. By setting x = a, we have c n = 1 n! f (n) (a). Thus, the Pre-U math is enough to obtain Taylor s expansion! f(x) = f(a) + f (a) 1! = j=0 (x a) + f (a) 2! (x a) 2 + f (a) (x a) 3 + 3! f (j) (a) (x a) j (2) n! Christopher Ting QF 101 Week 10 October 21, /70

14 A Primer of Partial Differentiation Bivariate function: f(x, y) Example: f(x, y) = x 3 + 5xy + y 2. Holding y constant, partial differentiation with respect to x: f(x, y) x = 3x 2 + 5y. Christopher Ting QF 101 Week 10 October 21, /70

15 A Primer of Partial Differentiation (Cont d) Holding x constant, partial differentiation with respect to y: f(x, y) y = 5x + 2y. Second-order partial derivatives: 2 f(x, y) x 2 = 6x, 2 f(x, y) y 2 = 2, 2 f(x, y) x y = 5. In general, 2 f(x, y) x y may not equal 2 f(x, y) y x. Christopher Ting QF 101 Week 10 October 21, /70

16 Bivariate Taylor s Expansion Recall pre-u s quadratic expansion (a + b) 2 = a 2 + 2ab + b 2. Let a := (x x 0 ) x and b := (y y 0) y. Then, for a bivariate function for which the order of partial differentiation is inter-changeable, ( (x x 0 ) x + (y y 0) ) 2 f(x, y) = (x x 0 ) 2 2 f y x 2 + 2(x x 0 )(y y 0 ) f y x + (y y 0) 2 2 f y 2 Christopher Ting QF 101 Week 10 October 21, /70

17 Bivariate Taylor s Expansion (Cont d) For bivariate function f(x, y) with respect to (x 0, y 0 ), f(x, y) = n=0 [( 1 (x x 0 ) n! x + (y y 0) ) n f(x, y)] y (x 0,y 0 ) Expanding up to the second order, we have f(x, y) f(x 0, y 0 ) + f x (x x 0) + f y (y y 0) + 1 [ 2 f 2! x 2 (x x 0) f x y (x x 0)(y y 0 ) + 2 f y 2 (y y 0) 2 ] Christopher Ting QF 101 Week 10 October 21, /70

18 Bivariate Taylor s Expansion (Cont d) Suppose x 0 = x dx, i.e., x x 0 = dx, an infinitesimally small difference from x 0. Likewise, y y 0 = dy. Moreover, df := f(x, y) f(x 0, y 0 ). Taylor s expansion in this case is, up to second order, df(x, y) f f dx + x y dy f 2 x 2 (dx)2 + 2 f x y dxdy f 2 y 2 (dy)2 (3) Christopher Ting QF 101 Week 10 October 21, /70

19 Einstein s Framework n particles are suspending in a liquid. Over a short period of time τ, the x coordinate of a particle will be displaced by a random amount ɛ, either positive or negative. Consider next the infinitesimal number dn of particles experiencing a displacement that lies between ɛ and ɛ + dɛ in the time interval τ. It is postulated that dn = nφ(ɛ)dɛ, where φ(ɛ) is a probability density function of ɛ. The probability for a particle to be found in the interval between ɛ and ɛ + dɛ is φ(ɛ)dɛ. For n particles, dn is therefore n times this probability. Christopher Ting QF 101 Week 10 October 21, /70

20 Statistical Dynamics Number of particles per unit volume: g(x, t) What is the distribution of the particles at a time t + τ, i.e., g(x, t + τ), from the distribution at time t, i.e., g(x, t)? Answer: g(x, ) is the sum of all possible values of g(x + ɛ, ) weighted by the probability φ(ɛ)dɛ. ( g(x, t + τ) dx = g(x + ɛ, t + τ)φ(ɛ)dɛ Now, since τ is very small, up to the first order g(x, t + τ) g(x, t) + τ g(x). t ) dx. (4) Christopher Ting QF 101 Week 10 October 21, /70

21 Einstein s Application of Taylor s Expansion Symmetric assumption, φ(ɛ) = φ( ɛ), necessitates ɛφ(ɛ) dɛ = 0. Since ɛ 1, i.e., very small, up to the second order, g(x, t) g(x + ɛ, t + τ)φ(ɛ) dɛ = φ(ɛ) dɛ + g ɛφ(ɛ) dɛ + 2 g ɛ 2 x x 2 φ(ɛ) dɛ + 2 g(x, t) + 2 g ɛ 2 x 2 φ(ɛ) dɛ. 2 Christopher Ting QF 101 Week 10 October 21, /70

22 Einstein s Coefficient of Diffusion Setting second-order moment C d := 1 2τ ɛ 2 φ(ɛ)dɛ, (5) and by virtue of (4), we obtain the well-known partial differential equation of diffusion, also known as the heat equation: g t = C 2 g d x 2. (6) In light of this equation, C d is identified as the coefficient of diffusion. Christopher Ting QF 101 Week 10 October 21, /70

23 Einstein s Solution of the Heat Equation Solution of (partial) differential equation needs the initial conditions. Suppose the motion of each particle starts at the point x for which the x-coordinate coincides with the center of gravity of n particles. For x 0 and t = 0, the initial conditions are The solution is g(x, 0) = 0, g(x, t) = and g(x, 0) dx = n. ) exp ( x2 n 4C d t. (7) 4πCd t Christopher Ting QF 101 Week 10 October 21, /70

24 Einstein s Scaling Law With g(x, t) (7), the variance of the displacement from the initial position x is x 2 g(x, t)dx = 2C d t. (8) This derivation requires the fact that exp ( x2 2 From (8), the root mean square displacement is proportional to the square root of time. x 2 g(x, t)dx t. ) dx = 2π. Christopher Ting QF 101 Week 10 October 21, /70

25 Bachelier s Framework The true price of an asset is at the origin, i.e., 0. The deviation or spread from the true price can in principle ranges from to. This difference from the true price is called the market price. Let p x,t1 dx denote the probability that, at time t 1 from time 0, the market price x is to be found in the infinitesimal interval ( x, x + dx ]. Similarly, p z x,t2 dz is the probability that the market price is at ( z x, z x + dz dx ]. Christopher Ting QF 101 Week 10 October 21, /70

26 Bachelier s Q&A What is the probability that the market price is in ( z, z + dz ] at time t 1 + t 2? If the price movement is independent, the said probability must be the product of these two probabilities: p x,t1 p z x,t2 dxdz. At time t 1, the market price could be located in any of the infinitesimal interval dx between and. The probability p z,t1 +t 2 dz of the price z being quoted at time t 1 + t 2 is Therefore, p z,t1 +t 2 dz = p z,t1 +t 2 = p x,t1 p z x,t2 dxdz. p x,t1 p z x,t2 dx. (9) Christopher Ting QF 101 Week 10 October 21, /70

27 Bachelier s Postulate With A and B denoting two constants with respect to x (but not time t), the solution for (9) is postulated to be p x, = A 2π exp ( B2 x 2 2 ). Being a probability function, it must be that 1 = p x, dx = A exp ( B2 x 2 ) dx. 2π 2 As a result, A = B. Christopher Ting QF 101 Week 10 October 21, /70

28 Boundary Condition Boundary condition: When the market price is equal to the true price, x = 0, we have p 0 = A 2π or B = A = p 0 2π. Therefore, the probability function becomes p x, = p 0 exp ( πp 2 0x 2). (10) Christopher Ting QF 101 Week 10 October 21, /70

29 Integration Let p 1 and p 2 be the quantities corresponding to p 0 but with respect to times t 1 and t 2, respectively. We evaluate (9) in light of (10): p z,t1 +t 2 = p 1 exp ( πp 2 1x 2) p 2 exp ( πp 2 2(z x) 2) dx. Noting that z is a constant in the integral, we have p z,t1 +t 2 =p 1 p 2 exp ( πp 2 2z 2) exp ( π ( p p 2 2) x 2 + 2πp 2 2z x ) dx. Christopher Ting QF 101 Week 10 October 21, /70

30 Completion of Square By completing the square, we write p z,t1 +t 2 = p 1p 2 exp ( πp 22z ) 2 + πp4 2 z2 p p 2 p e πu2 du, p2 2 where the variable has been transformed to u := x p p2 2 p 2 2 z. p p 2 2 Since the integral p z,t1 +t 2 = p 1p 2 p p 2 2 e πu2 du = 1, we obtain ( ) exp π p2 1 p2 2 p z 2. p2 2 Christopher Ting QF 101 Week 10 October 21, /70

31 Bachelier s Solution If we write t 3 = t 1 + t 2 and define p 3 := p 1p 2, (11) p p 2 2 then the same functional form (10) for the probability function is evident: p z,t3 = p 3 exp ( πp 2 3 z 2). Hence, the solution for (9) is obtained. Christopher Ting QF 101 Week 10 October 21, /70

32 Nice Math! Going further, we define, for i = 1, 2, the function f(t i ) as follows: p i := f(t i). 2π Inspired by (11) and squaring both sides, we obtain f 2 (t 1 + t 2 ) = f 2 (t 1 )f 2 (t 2 ) f 2 (t 1 ) + f 2 (t 2 ). The function f(t 1 + f 2 ) is symmetric in t 1 and t 2. Partial differentiation with respect to t 1 is therefore equal to the partial differentiation with respect to t 2. Accordingly, f (t 1 ) f 3 (t 1 ) = f (t 2 ) f 3 (t 2 ). Christopher Ting QF 101 Week 10 October 21, /70

33 Final Solution We see the separation of variables in this equation and thus the ratio is a constant C. If follows that f (t) = Cf 3 (t). (12) It is easy to verify that the solution of (12) is f(t) = 2π p 0 = D t, where D is a constant. In accordance to (10), the probability density function is p(x, t) = D exp ( D2 x 2 ). (13) 2πt 2t Christopher Ting QF 101 Week 10 October 21, /70

34 Bachelier s Scaling Law The expected value of x 2 i.e., variance σ 2 (t) is found to be σ 2 (t) = x 2 D exp ( D2 x 2 ) dx = t 2π t 2t D 2. The variance σ 2 (t) is proportionate to time t. In other words, the square root of the mean of squared displacement from 0 is proportional to the square root of elapsed time σ(t) := 1 D t t. This scaling law is exactly the same as Einstein s scaling law (8)! Christopher Ting QF 101 Week 10 October 21, /70

35 Link to Finance For a unit time t = 1, we define σ 2 := σ 2 (1) as the rate of variance. The constant D is thus given by D = 1 σ. Consequently, the probability function is p(x, t) = ) 1 exp ( x2 2πt σ 2σ 2. (14) t Volatility σ, le coefficient d instabilité, of return (not the price!) is a proxy for (potential) risk. Christopher Ting QF 101 Week 10 October 21, /70

36 Mathematical Definition of Brownian Motion Let t 0 denote time. A stochastic process B t is said to be Brownian when (1) B 0 = 0; (2) {B t, t 0} has stationary independent increments; (3) for every t > 0, B t is normally distributed with mean 0 and variance proportional to t. The first condition sets the origin of the coordinate system. The third condition is the scaling law of diffusion. Christopher Ting QF 101 Week 10 October 21, /70

37 Intuition Underlying the Second Condition The second condition of stationarity is a statement about the invariance of the system with respect to a translation in the time coordinate. Namely, having the same statistical distribution, the increment B t := B(t + dt) B t, as a random variable, is the same as B(t + s) for all t. Also, the system has zero memory since each B t is independently distributed. Consequently, the correlation of B t and B(t + s) vanishes when s 0. Christopher Ting QF 101 Week 10 October 21, /70

38 Infinitesimal Increment Suppose we start the Brownian motion process at time t = 0 and consider an infinitesimal increment ɛ defined as ɛ := B = B( t) B(0) = B( t) after an infinitesimal time t has elapsed. From the third defining condition, the variance of the infinitesimal increment is normally distributed with mean 0, and with an appropriate choice of scale, the variance of ɛ is σ 2 t: V ( ɛ ) σ 2 t, which is the same as saying that E ( ɛ 2) = σ 2 t. (15) Christopher Ting QF 101 Week 10 October 21, /70

39 Variance of ( B) 2 Equation (15) suggests that the expected value of ( B) 2 is σ 2 t. We want to show that, up to the first order in t, when σ = 1. lim t 0 ( B)2 (db) 2 dt, (16) This is an important component of Itô s formula in differential form. Equation (16) says that the square of the random variable db is deterministic! To prove equation (16), we shall show that the variance of ( B) 2 is indeed very small, and vanishes when t is an infinitesimal. Christopher Ting QF 101 Week 10 October 21, /70

40 Application of Pre-U Integration by Parts We start from Equation (15). Knowing that the probability distribution function is normal, specifically N ( 0, σ 2 t ), we can write it as σ 2 t = 1 2π t σ ) x 2 exp ( x2 2σ 2 dx. t Regarding the integrand ) as made up of two functions, x and x exp ( x2 2σ 2, we perform the integration by parts. t Christopher Ting QF 101 Week 10 October 21, /70

41 Appearance of Kurtosis We thus obtain ( σ 2 1 x 2 t = 2π t σ 2 x exp x 2 ( exp 2 ( x2 2σ 2 t ( x2 ) 2σ 2 t - ) x2 σ 2 t exp The first term vanishes because its exponential term vanishes at a much faster rate than x 3 going towards infinity. Therefore, we have σ 2 1 t = 2π t σ + 1 2π t σ x 2 ( 2 exp x2 x 4 2σ 2 t exp 2σ 2 t )) ) ( x2 2σ 2 dx. t ) dx ( x2 2σ 2 t ) dx. (17) Christopher Ting QF 101 Week 10 October 21, /70

42 Analytical Expression of Kurtosis Note that the first term of Equation (17) is equal to σ 2 t/2. Re-arranging the terms, we obtain 3 2 σ2 t = 1 2π t σ x 4 ) ( 2σ 2 t exp x2 2σ 2 dx. t In other words, we have obtained an analytical expression for the integral 1 2π t σ ) x 4 exp ( x2 2σ 2 dx = 3σ 4( t ) 2. (18) t Christopher Ting QF 101 Week 10 October 21, /70

43 Variance of Variance With this result, it becomes straightforward to compute the variance of ɛ 2 : V ( ɛ 2) = E ( (ɛ 2 σ 2 t ) 2 ) = E ( ɛ 4) σ 4( t ) 2. (19) From equation (18), we obtain the result E ( ɛ 4) = 3σ 4 ( t) 2, i.e., the kurtosis of a normal distribution is 3, E ( ɛ 4) /E ( ɛ 2) 2 = 3. It follows that the variance of ɛ 2 as defined by equation (19) is equal to 2σ 4 ( t) 2, i.e., V ( ɛ 2) = 2σ 4 ( t) 2. Christopher Ting QF 101 Week 10 October 21, /70

44 From Deterministic to Stochastic Since ɛ = B, we see that the variance of ( B) 2 is 2σ 4 ( t) 2, which is negligibly small compared to t when t 0. It is tantamount to saying that ɛ 2 does not vary stochastically, for its variance is almost zero. Therefore, we can conclude that, with σ set equal to 1 and in the ultimate limit when t 0, and from equation (15), ( ( B ) ) 2 E ( B) 2 t. Summary: While B N(0, t) is a random variable, its square is informally written as (db) 2 dt. Christopher Ting QF 101 Week 10 October 21, /70

45 Symbolic Rules of Itô Calculus Heuristically speaking, db dt. With respect to dt, the product db dt is of a higher order and hence equals to 0. In the infinitesimal limit t 0, we have the following symbolic rules to perform stochastic calculus: db t dt db t dt 0 dt 0 0 Christopher Ting QF 101 Week 10 October 21, /70

46 Itô Process and Itô Formula Consider a stochastic process X(t) consisting of two parts a deterministic part and a stochastic part: dx(t) = g(t) dt + σ(t) db t. (20) This is the stochastic differential equation (SDE) for X(t), and is named the Itô process. Suppose f(t, x) is a continuous function with continuous partial derivatives f t, f x, and 2 f x 2. Then f ( t, X(t) ) is also an Itô process. Moreover, its SDE is df ( t, X(t) ) ( f t + f x g(t) ) f 2 x 2 σ(t)2 dt+ f x σ(t) db t. (21) Christopher Ting QF 101 Week 10 October 21, /70

47 Symbolic Proof Apply the 2-d Taylor expansion (3) to obtain, up to dt, df f f dt + t x dx(t) f ( ) 2 dx(t). (22) 2 x 2 Then use the symbolic table to get ( ) 2 ( ) 2 dx(t) = g(t) dt + σ(t) dbt Therefore, df f t = ( g(t) dt ) 2 + 2g(t)f(t)dBt dt + ( σ(t) db t ) 2 σ(t) 2 dt. f ) dt + (g(t) dt + σ(t) db t f x 2 ( f t + f x g(t) f 2 x 2 σ(t)2 x 2 σ(t)2 dt ) dt + f x σ(t) db t. Christopher Ting QF 101 Week 10 October 21, /70

48 Itô Calculus What Kiyoshi Itô did was to show that is really =. This is indeed a great achievement. Christopher Ting QF 101 Week 10 October 21, /70

49 Differential Equations Determining a Markov Process Zenkoku Sizyo Sugaku Danwakai (1942) (Journal of Pan-Japan Mathematical Colloquium) No.1077, Source of the original paper: Zenkoku Sizyo Sugaku Danwakai, Osaka University Source: Christopher Ting QF 101 Week 10 October 21, /70

50 Itô s 1942 Paper Let f(x) be a twice differentiable function, not many people know that Itô s formula in its rudimentary form had already appeared: t 0 f(x t )dx t = Xt Shifting the terms around, Xt 0 f(λ)dλ = t 0 f(λ)dλ 1 2 f (X t )dt + t t 0 0 f (X t )dt. f(x t )dx t. Differentiation with respect to x t gives rise to the SDE df(x t ) xt=x t = 1 2 f xt=x t dt + f (x t ) xt=x t dx t. Christopher Ting QF 101 Week 10 October 21, /70

51 Itô s 1942 Paper (Cont d) We write the rudimentary form of Itô s formula as df(x t ) = 1 2 f (X t )dt + f (X t )dx t. (23) Remarks: Brownian motion is also called Wiener s process. A stochastic sample path X t, though continuous, Norbert Wiener (1923) shows that it is no where differentiable. That s (one of the reasons) why Itô thought of integrating a random path but not to differentiate it. Though the sample path X t is non-differentiable, the function f(x t ) can be differentiated by the usual Leibniz Newton calculus. SDE is purely a symbolic way of abbreviating Itô s integral. Many textbooks still call it Itô s lemma. But this is a disrespect of Itô s momentous contribution to math. Christopher Ting QF 101 Week 10 October 21, /70

52 First Example of Itô s Calculus In 7 (Page 1374), as the first example of stochastic integration, t 0 B s db s = 1 2 B2 t 1 2 t. Proof: Let f(t, x) = 1 2 x2. Then we have f x = x, 2 f x 2 = 1. By applying the Itô formula, equation (23), we obtain df(t) = 1 2 dt + B tdb t. Christopher Ting QF 101 Week 10 October 21, /70

53 First Example of Itô s Calculus (Cont d) Taking the integral, the result is t 0 df(s) = 1 2 t 0 ds + t 0 B s db s, Since the functional form of f( ) is 1 2 x2, we get 1 2 B2 t = 1 2 t + t 0 B s db s. Christopher Ting QF 101 Week 10 October 21, /70

54 Very Important Model in Quantitative Finance Geometric Brownian motion S t, is a positive definite stochastic process ds t = µ(t)s t dt + σ(t)s t db t, which is more appropriately written instead as Log price log ( S t ) is a function of St. ds t S t = µ(t) dt + σ(t) db t. (24) Now, f(t, x) = log x, we have f x = 1 x and 2 f x 2 = 1 x 2. Christopher Ting QF 101 Week 10 October 21, /70

55 Very Important Model in Quantitative Finance (Cont d) Applying the Itô 1942 formula (23), we get d ( ) 1 log S t = ds t 1 1 ( ) 2 S t 2 S 2 dst t = ds t 1 ( ) 2 dst S t 2 S t Now, the square of (24) is ( ) 2 dst = S 2 t σt 2 ( ) 2 dbt = S 2 t σt 2 dt, (25) Christopher Ting QF 101 Week 10 October 21, /70

56 Very Important Model in Quantitative Finance (Cont d) Substitute in equation (24) and apply the symbolic rules, d ( ) log S t = µ(t) dt + σ(t) dbt 1 2 σ2 (t) dt ( = µ(t) 1 ) 2 σ2 (t) dt + σ(t) db t. Integrating over time, we obtain ( t ( S t = S 0 exp µ(t) 1 ) 2 σ2 (t) dt + 0 t 0 σ(t) db t ). Christopher Ting QF 101 Week 10 October 21, /70

57 Very Important Model in Quantitative Finance (Cont d) If µ(t) and σ(t) are constants, then since B(0) = 0, S t = S 0 exp ((µ 12 ) ) σ2 t + σb t. (26) This geometric Brownian motion (26) with constant µ and σ is an important model of asset price. Christopher Ting QF 101 Week 10 October 21, /70

58 The Black-Scholes Equation Assume that the underlying asset price follows (26). Construct a portfolio P t comprising the option O(t, S t ) and δ units of the underlying asset S t. P t = O(t, S t ) + δs t. (27) The brilliant idea of Black and Scholes (1973) is that, by adjusting δ, the portfolio can be made risk-free. By the first principle of Quantitative Finance, any risk-free asset must increase at the risk-free rate r. If the principal sum P t is deposited at the risk-free bank, then P t r t is the interest earned over a short time period t. Christopher Ting QF 101 Week 10 October 21, /70

59 The Black-Scholes Equation (Cont d) In view of (27), the change in portfolio must be, with the abbreviation O t := O(t, S t ), In the limit t 0, we have P t = O t + δ S t = P t r t. (28) dp t = do t + δ ds t = r ( O t + δ S t ) dt. Since the option O := O(t, S t ) is a function of the stochastic process S t, Black and Scholes applied the Itô formula (22) to obtain ( ) O O t dt + + δ ds t O ( ) 2 ( ) S t 2 St 2 dst = r O + δst dt. Christopher Ting QF 101 Week 10 October 21, /70

60 The Black-Scholes Equation (Cont d) Equivalently, in view of (25), ( O t σ2 St 2 2 ) ( ) O O St 2 dt + + δ ds t = r ( ) O + δs t dt. S t (29) To make the portfolio risk-free, you just need to remove the random term ds t away by setting the hedge ratio δ to δ = O S t. (30) Substituting this δ into (29), you obtain the Black-Scholes equation for pricing option: O t + rs O t + 1 S t 2 σ2 St 2 2 O S 2 t r O = 0. (31) Christopher Ting QF 101 Week 10 October 21, /70

61 17 Equations that Changed the World Here are some traders in the S&P 500 options pit at the CBOE. You won t find a single person here that hasn t heard about the Black-Scholes equation. Christopher Ting QF 101 Week 10 October 21, /70

62 Black-Scholes Equation = Heat Equation The Black-Scholes equation can be transformed into the heat equation when the underlying price S t is transformed into z = log(s t ) + (r 12 ) σ2 (T t). The heat equation is where τ := T t, and u = e rt O. What is the meaning of τ? Answer: What is the meaning of u? Answer: What is the meaning of σ 2 /2? Answer: u τ = 1 2 σ2 2 u z 2, (32) Christopher Ting QF 101 Week 10 October 21, /70

63 Black-Scholes Option Pricing Formulas d 1 and d 2 d1 = log ( ) ( S t K + r σ2) τ σ, τ d 2 = log ( ) ( S t K + r 1 2 σ2) τ σ, τ Standard normal cumulative distribution function: Φ(x) := 1 x e ( ) v2 2 dv =: P X x 2π The Black-Scholes pricing formulas for European calls and puts c(t, S t ) = S t Φ(d 1 ) Ke r(t t) Φ(d 2 ) p(t, S t ) = Ke r(t t) Φ( d 2 ) S t Φ( d 1 ) Christopher Ting QF 101 Week 10 October 21, /70

64 Experience of a Quantitative Option Trader I didn t see how to prove the formula but I decided to go ahead and use it to invest, because there was in an abundance of vastly overpriced (in the sense of Beat the Market) OTC options.... The formula has proven itself in action. Edward Thorp Source: Option Theory What I knew and When I Knew It Christopher Ting QF 101 Week 10 October 21, /70

65 Takeaways Brownian motion is a continuous-time limit of random walk. The generalization of Maclaurin series is the Taylor expansion series (the Taylor theorem). Partial differentials are simple extension of single-variable differentials to multi-variable cases. Einstein applied the Taylor theorem to obtain the scaling law of Brownian diffusion. Bachelier used clever pre-u math to derive the scaling law for random variables associated to the Gaussian distribution. Symbolic rules of Itô s calculus are useful in analyzing stochastic differential equation. Christopher Ting QF 101 Week 10 October 21, /70

66 Takeaways (Cont d) The symbolic proof of Itô s formula hinges upon the Taylor theorem. The derivation of geometric Brownian motion is to be done in Itô s calculus. The key of the Black and Scholes method is the idea of delta-hedging away the uncertainty due to the fluctuation of the underlying asset price. In the derivation of their equation. Black and Scholes are the first ever to apply the Itô formula. The Black and Scholes equation is mathematically equivalent to the heat equation. The Black and Scholes formulas for pricing European options are the continuous-time limit of the binomial models. Christopher Ting QF 101 Week 10 October 21, /70

67 Week 10 Assignment 1. Exercises for Taylor s expansion up to third-order differential (i.e., 4 terms). (A) e x about a = 0 (B) a x about a = 1 (C) 1 x about a = 1 (D) log(x) about a = 2 (E) 1 about a = 1 x2 (F) cos(x) about a = π 2. Show that the call and put formulas in Slide 63 satisfy the put-call parity. Christopher Ting QF 101 Week 10 October 21, /70

68 Week 10 Assignment (Cont d) 3. Q3 of Chapter 8 4. Q4 of Chapter 8 5. Q5 of Chapter 8 Christopher Ting QF 101 Week 10 October 21, /70

69 Week 10 Additional Exercise 1 Log return is also called continuously compounded return ( ) Pt r t := log(1 + R t ) = log = p t p t 1 P t 1 Let D t be the stock dividend in dollars and define the log dividend yield: δ t := log(d t ) log(p t ) := d t p t. (A) Define the log return inclusive of dividend h t := log ( ) P t + D t log(pt ). Show that h t is a function of δ t : h t (δ t ) := p t p t 1 + log ( 1 + e δt). Christopher Ting QF 101 Week 10 October 21, /70

70 Week 10 Additional Exercise (Cont d) (B) Apply Taylor s expansion of h t (δ t ) about the average log dividend yield δ and show that where h t (δ t ) p t p t 1 + (1 ρ)δ t log(ρ) (1 ρ) δ, ρ := e δ (C) Define k := log(ρ) (1 ρ) δ. Demonstrate and interpret (D) Suppose h t (δ t ) = ρp t p t 1 + (1 ρ)d t + k. Demonstrate and interpret j=0 lim j ρj p t+j = 0. p t ρ j (1 ρ)d t+1+j ρ j h t+1+j + k 1 ρ. (33) j=0 Christopher Ting QF 101 Week 10 October 21, /70